Answer:
For the combination A, then B, then AB => 0.164%
For any combination (incl. B, AB, A or A, AB, B) => 0.984%
Step-by-step explanation:
You're looking for any combination of students which have the blood types A, B, AB. Any single one combination will have a probability that looks like the following:
Probability of A * Probability of B * Probability of AB
Let's convert the probabilities to decimal.
P(A) = 41/100 = 0.41
P(B) = 1/10 = 0.10
P(AB) = 1/25 = 0.04
Any ONE combination, e.g. (B, AB, A) or (A, AB, B), will thus have the probability
0.41 * 0.10 * 0.04 = 0.00164
Note: the following ONLY applies if you want to find the probability of ANY combination of A, B, AB. If you only want the one case where the first random student is A, the 2nd is B, the 3rd is AB, then the answer is 0.00164 = 0.164%
If we want to find the probability of EVERY case combined that has A, B, AB -> we need to answer the question: how many combinations are there?
Student 1 can be any of the 3 blood types.
For each of the blood types that student 1 can be, student 2 can be one of 2 other blood types.
Whatever the previous combination, student 3 can only be 1 blood type: the one remaining.
3 cases for Student 1, each case having 2 cases for Student 2, each having only 1 case for Student 3.
3 * 2 * 1 = 6
So 6 cases, where each probability is 0.00164
6 * 0.00164 = 0.00984 = 0.984%
